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P[email protected]·General Discussion· 9 views·2 weeks ago

The dispersion-limited bandwidth-distance product for a standard SMF at 1550 nm is approximately

In optical fiber transmission, 'Chromatic Dispersion' causes pulse broadening. The dispersion-limited
bandwidth-distance product for a standard SMF at 1550 nm is approximately?

what will be its value?

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Anonymous·2 weeks ago

ITU-T G.Sup39 (*Optical system design and engineering considerations*, https://www.itu.int/rec/T-REC-G.Sup39/en) frames this as a dispersion allowance derived from source spectral width, modulation format, and an allowed eye-closure penalty — not as a fixed fiber constant. Two criteria are in common use and they give different numbers, so state which one you mean.

**Eye-closure penalty form (G.Sup39):**

$$D \cdot L \cdot \sigma_\lambda \cdot B \leq \epsilon, \qquad \epsilon \approx 0.306\ (1\,\text{dB}), \quad \epsilon \approx 0.491\ (2\,\text{dB})$$

(Gaussian pulse, chirp-free source.)

**Broad source** ($\sigma_\lambda \approx 1$ nm, $D \approx 17$ ps/(nm·km) on G.652):

$$BL \leq \frac{\epsilon}{D\sigma_\lambda} = \frac{0.306}{17 \times 1} \approx 18\ \text{(Gb/s)}\cdot\text{km}$$

The simpler $4\sigma$ pulse-spread criterion gives a slightly tighter figure:

$$BL \leq \frac{1}{4D\sigma_\lambda} = \frac{1}{4 \times 17 \times 1} \approx 15\ \text{(Gb/s)}\cdot\text{km}$$

**Narrow-line DFB** — source width is set by the modulation itself, so the invariant is $B^2L$ rather than $BL$. With $\beta_2 = -\lambda^2 D / 2\pi c \approx -21.7$ ps²/km:

$$B^2 L \leq \frac{1}{16|\beta_2|} = \frac{1}{16 \times 21.7} \approx 3000\ \text{(Gb/s)}^2\cdot\text{km}$$

→ ≈ 30 km at 10 Gb/s, i.e. ~1 (Tb/s)·km equivalent at 100 km.

So: ≈ 15–18 (Gb/s)·km for the 1 nm-source case depending on criterion, ~1 (Tb/s)·km narrow-line. Either number is incomplete without the source width and penalty budget stated alongside it.

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